3.128 \(\int \sin ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\)

Optimal. Leaf size=159 \[ -\frac {\sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac {b}{a}\right .\right )}{3 b f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(a+2 b) \sqrt {a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{3 b f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}} \]

[Out]

-1/3*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/f+1/3*(a+2*b)*(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticE(si
n(f*x+e),(-b/a)^(1/2))*(a+b*sin(f*x+e)^2)^(1/2)/b/f/(1+b*sin(f*x+e)^2/a)^(1/2)-1/3*a*(a+b)*(cos(f*x+e)^2)^(1/2
)/cos(f*x+e)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^(1/2)/b/f/(a+b*sin(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3170, 3172, 3178, 3177, 3183, 3182} \[ -\frac {\sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {a (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac {b}{a}\right .\right )}{3 b f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(a+2 b) \sqrt {a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{3 b f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

-(Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*f) + ((a + 2*b)*EllipticE[e + f*x, -(b/a)]*Sqrt[a +
 b*Sin[e + f*x]^2])/(3*b*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) - (a*(a + b)*EllipticF[e + f*x, -(b/a)]*Sqrt[1 + (b
*Sin[e + f*x]^2)/a])/(3*b*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 3170

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[(B*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(2*f*(p + 1)), x] + Dist[1/(2*(p + 1)), Int[(a + b*Si
n[e + f*x]^2)^(p - 1)*Simp[a*B + 2*a*A*(p + 1) + (2*A*b*(p + 1) + B*(b + 2*a*p + 2*b*p))*Sin[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, e, f, A, B}, x] && GtQ[p, 0]

Rule 3172

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[
B/b, Int[Sqrt[a + b*Sin[e + f*x]^2], x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /;
FreeQ[{a, b, e, f, A, B}, x]

Rule 3177

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[e + f*x, -(b/a)])/f, x]
 /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3178

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[1 + (b*Sin
[e + f*x]^2)/a], Int[Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rule 3182

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1*EllipticF[e + f*x, -(b/a)])/(Sqrt[a]*
f), x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3183

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[1 + (b*Sin[e + f*x]^2)/a]/Sqrt[a +
b*Sin[e + f*x]^2], Int[1/Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \sin ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx &=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {1}{3} \int \frac {a+(a+2 b) \sin ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\\ &=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {(a (a+b)) \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx}{3 b}+\frac {(a+2 b) \int \sqrt {a+b \sin ^2(e+f x)} \, dx}{3 b}\\ &=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {\left ((a+2 b) \sqrt {a+b \sin ^2(e+f x)}\right ) \int \sqrt {1+\frac {b \sin ^2(e+f x)}{a}} \, dx}{3 b \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {\left (a (a+b) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \int \frac {1}{\sqrt {1+\frac {b \sin ^2(e+f x)}{a}}} \, dx}{3 b \sqrt {a+b \sin ^2(e+f x)}}\\ &=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {(a+2 b) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{3 b f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {a (a+b) F\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 b f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.83, size = 159, normalized size = 1.00 \[ \frac {b \sin (2 (e+f x)) (-2 a+b \cos (2 (e+f x))-b)-2 \sqrt {2} a (a+b) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )+2 \sqrt {2} a (a+2 b) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{6 \sqrt {2} b f \sqrt {2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(2*Sqrt[2]*a*(a + 2*b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] - 2*Sqrt[2]*a*(a + b)
*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + b*(-2*a - b + b*Cos[2*(e + f*x)])*Sin[2*(
e + f*x)])/(6*Sqrt[2]*b*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (\cos \left (f x + e\right )^{2} - 1\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-b*cos(f*x + e)^2 + a + b)*(cos(f*x + e)^2 - 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*sin(f*x + e)^2, x)

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maple [A]  time = 1.45, size = 266, normalized size = 1.67 \[ -\frac {-b^{2} \left (\sin ^{5}\left (f x +e \right )\right )+\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b -a b \left (\sin ^{3}\left (f x +e \right )\right )+b^{2} \left (\sin ^{3}\left (f x +e \right )\right )+a b \sin \left (f x +e \right )}{3 b \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2),x)

[Out]

-1/3*(-b^2*sin(f*x+e)^5+(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))
*a^2+a*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-(cos(f*x+e)^2)
^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2-2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(
f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b-a*b*sin(f*x+e)^3+b^2*sin(f*x+e)^3+a*b*sin(f*x+e))/
b/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*sin(f*x + e)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^2\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^2*(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(sin(e + f*x)^2*(a + b*sin(e + f*x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**2*(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(e + f*x)**2)*sin(e + f*x)**2, x)

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